iCON Education - Some Basic Concepts of Chemistry

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Online) 9th April Morning Slot

What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution ?

A.

320

B.

325

C.

316

D.

330

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Online) 8th April Morning Slot

Excess of NaOH (aq) was added to 100 mL of FeCl₃ (aq) resulting into 2.14 g of Fe(OH)₃ . The molarity of FeCl₃ (aq) is :

(Given molar mass of Fe = 56 g mol⁻¹ and molar mass of Cl = 35.5 g mol₋₁)

A.

0.2 M

B.

03 M

C.

0.6 M

D.

1.8 M

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Offline)

The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all ¹H atoms are replaced by ²H atoms is:

A.

37.5 kg

B.

7.5 kg

C.

10 kg

D.

15 kg

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Offline)

1 gram of a carbonate (M₂CO₃) on treatment with excess HCl produces 0.01186 mole of CO₂. The molar mass of M₂CO₃ in g mol^–1 is:

A.

84.3

B.

118.6

C.

11.86

D.

1186

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Online) 9th April Morning Slot

An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% sulphur is :

(atomic weight of S = 32 amu)

A.

200 g mol^$-$1

B.

400 g mol^$-$1

C.

600 g mol⁻¹

D.

300 g mol⁻¹

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Online) 9th April Morning Slot

The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H₂S in the presence of conc. HCl ( assuming 100% conversion) is :

A.

0.50 mol

B.

0.25 mol

C.

0.125 mol

D.

0.333 mol

2015 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2015 (Offline)

The molecular formula of a commercial resin used for exchanging ions in water softening is C₈H₇SO₃Na (Mol. Wt. 206). What would be the maximum uptake of Ca²⁺ ions by the resin when expressed in mole per gram resin?

A.

2/309

B.

1/412

C.

1/103

D.

1/206

2014 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2014 (Offline)

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is:

A.

1: 8

B.

3 : 16

C.

1 : 4

D.

7 : 32

2014 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2014 Paper 1 Offline

If the value of Avogadro number is 6.023 $\times$ 10²³ mol^-1 and the value of Boltzmann constant is 1.380 $\times$ 10^-23 J K^-1, then the number of significant digits in the calculated value of the universal gas constant is

2014 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2014 Paper 1 Offline

A compound H₂X with molar weight of 80g is dissolved in a solvent having density of 0.4 gml^–1 . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

2013 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2013 (Offline)

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO₂. The empirical formula of the hydrocarbon is

A.

C₂H₄

B.

C₃H₄

C.

C₆H₅

D.

C₇H₈

2013 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2013 (Offline)

Experimentally it was found that a metal oxide has formula M_0.98O. Metal M, present as M²⁺ and M³⁺ in its oxide. Fraction of the metal which exists as M³⁺ would be

A.

7.01%

B.

4.08%

C.

6.05%

D.

5.08%

2013 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2013 (Offline)

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:

A.

1.00 M

B.

1. 75 M

C.

0.975 M

D.

0.875 M

2012 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2012

The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL. The molarity of this solution is :

A.

0.50 M

B.

1.78 M

C.

1.02 M

D.

2.05 M

2012 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2012 Paper 1 Offline

29.2 % (w/w) HCl stock solution has density of 1.25 g mL^-1 . The molecular weight of HCl is 36.5 g mol^-1 . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

Correct Answer: 8

Explanation:

To calculate the volume of the 29.2% (w/w) HCl stock solution needed to prepare a 200 mL solution of 0.4 M HCl, we need to use several steps involving concentration and density conversions.

First, we calculate the mass of HCl that is contained in the 200 mL of a 0.4 M solution:

$ Mass = Molarity \times Volume \times Molecular\ Weight $

$ Mass = 0.4 \ mol/L \times 0.200 \ L \times 36.5 \ g/mol $

Note that we convert the volume from mL to L to match the units of molarity (mol/L).

Now, we calculate it:

$ Mass = (0.4 \times 0.200 \times 36.5) \ g $

$ Mass = 0.08 \times 36.5 \ g $

$ Mass = 2.92 \ g $

The next step is to determine how much of the stock solution is needed to get 2.92 g of HCl. Since the stock solution is 29.2% (w/w) HCl, this means that in every 100 g of stock solution, there is 29.2 g of HCl. We can set up a proportion to find the mass of the stock solution needed:

$ \frac{29.2\ g \ HCl}{100\ g \ stock\ solution} = \frac{2.92\ g \ HCl}{x\ g \ stock\ solution} $

Now we solve for $ x $:

$ x = \frac{2.92\ g \times 100\ g \ stock\ solution}{29.2\ g \ HCl} $

$ x = \frac{292}{29.2} \ g $

$ x = 10\ g $

So, we need 10 g of the stock solution to get 2.92 g of HCl.

The final step is to calculate the volume of the stock solution that has a mass of 10 g. We use the density to convert mass to volume:

$ Volume = \frac{Mass}{Density} $

The density of the stock solution is given as 1.25 g/mL, so:

$ Volume = \frac{10\ g}{1.25\ g/mL} $

$ Volume = 8\ mL $

Therefore, to prepare a 200 mL solution of 0.4 M HCl, you would need to measure out 8 mL of the 29.2% HCl stock solution.

2011 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2011

A 5.2 molal aqueous solution of methyl alcohol, CH₃OH, is supplied. What is the mole fraction of methyl alcohol in the solution ?

A.

0.190

B.

0.086

C.

0.050

D.

0.100

2011 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2011 Paper 2 Offline

The volume (in mL) of 0.1 M AgNO₃ required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of $[Cr{({H_2}O)_5}Cl]C{l_2}$, as silver chloride is close to ____________.

2011 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2011 Paper 1 Offline

Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is

A.

1.78 M

B.

2.00 M

C.

2.05 M

D.

2.22 M

Correct Answer: C

Explanation:

To find the molarity of the urea solution, we will follow these steps:

1. Calculate the number of moles of urea using its molecular weight.

2. Determine the volume of the solution using its mass and density.

3. Use the number of moles and volume of the solution to calculate the molarity.

Step 1: Calculate the number of moles of urea.

The number of moles of a substance $ n $ is determined by dividing the mass of the substance ($ m $) by its molecular weight ($ M $):

$ n = \frac{m}{M} $

For urea, $ m = 120 \text{ g} $ and $ M = 60 \text{ g/mol} $, thus:

$ n_{urea} = \frac{120 \text{ g}}{60 \text{ g/mol}} $

$ n_{urea} = 2 \text{ moles} $

Step 2: Determine the volume of the solution.

The volume $ V $ of the solution can be found by dividing the mass of the solution by its density $ \rho $:

$ V = \frac{mass_{solution}}{\rho} $

The mass of the solution is the sum of the mass of urea and the mass of water. Thus:

$ mass_{solution} = mass_{urea} + mass_{water} = 120 \text{ g} + 1000 \text{ g} $

$ mass_{solution} = 1120 \text{ g} $

The density of the solution $ \rho $ is given as 1.15 g/mL. Therefore, the volume in milliliters (which is equivalent to cubic centimeters) is:

$ V = \frac{1120 \text{ g}}{1.15 \text{ g/mL}} $

$ V = 973.91 \text{ mL} $

To convert milliliters to liters (since molarity is defined in terms of liters), we divide by 1000:

$ V = 0.97391 \text{ L} $

Step 3: Calculate the molarity.

Molarity ($ M $) is defined as the number of moles of solute divided by the volume of solution in liters:

$ M = \frac{n_{solute}}{V_{solution}} $

$ M = \frac{2 \text{ moles}}{0.97391 \text{ L}} $

$ M \approx 2.05 \text{ M} $

Therefore, the molarity of the urea solution is approximately 2.05 M, which corresponds to Option C.

2010 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2010 Paper 1 Offline

A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is

Correct Answer: 3

Explanation:

To find the average titre value, first add up the three measurements provided and then divide by the number of measurements.

The sum of the measurements is:

$ 25.2 \, \text{mL} + 25.25 \, \text{mL} + 25.0 \, \text{mL} = 75.45 \, \text{mL} $

Since there are three measurements, divide this sum by 3 to calculate the average:

$ \text{Average titre value} = \frac{75.45 \, \text{mL}}{3} = 25.15 \, \text{mL} $

When reporting the average, we must consider the significant figures of the original measurements. The number of significant figures is determined by the least precise measurement, which in this case is 25.0 mL with three significant figures. Therefore, we should report the average value to three significant figures as well.

The average value of 25.15 mL has four significant figures, so we need to round it to three significant figures. However, this is slightly tricky since 25.15 already appears to be rounded to four significant figures. We should consult the original measurements to decide on the best course of action.

Looking at the individual measurements (25.2, 25.25, and 25.0), we should consider the lowest decimal place which they all have in common, which is the first decimal place. The third measurement has no second decimal place, indicating its level of precision. Thus, the number of significant figures for the average titre value should be in line with this level of precision. Since the average calculated is 25.15, when we adjust to the first decimal place for consistent significant figures, the average is 25.1 mL with three significant figures.

$ \text{Corrected Average titre value} = 25.1 \, \text{mL} $

Therefore, the number of significant figures in the average titre value is three: 25.1 mL.

2010 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2010 Paper 2 Offline

Silver (atomic weight = 108 g mol^-1) has a density of 10.5 g.cm^-3. The number of silver atoms on a surface of area 10^-12 m² can be expressed in scientific notation as y $\times$ 10^x. The value of x is?

2009 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2009 Paper 1 Offline

Given that the abundances of isotopes ⁵⁴Fe, ⁵⁶Fe and ⁵⁷Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is :

A.

55.85

B.

55.95

C.

55.75

D.

56.05

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

The density (in g mL^-1) of a 3.60 M sulphuric acid solution that is 29% H₂SO₄ (molar mass = 98 g mol^-1) by mass will be

A.

1.45

B.

1.64

C.

1.88

D.

1.22

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

In the reaction

2Al(s) + 6HCl(aq) $\to$ 2Al³⁺ (aq) + 6Cl^-(aq) + 3H₂(g)

A.

11.2 L H₂(g) at STP is produced for every mole HCl(aq) consumed

B.

6 L HCl(aq) is consumed for every 3L H₂(g) produced

C.

33.6 L H₂(g) is produced regardless of temperature and pressure for every mole that reacts

D.

67.2 H₂(g) at STP is produced for every mole Al that reacts

Correct Answer: A

Explanation:

Option A :
6 moles of HCl(aq) produces = 3 moles of H₂(g)

$\therefore$ 1 mole of HCl(aq) produces = ${3 \over 6}$ moles of H₂(g) = ${1 \over 2}$ moles of H₂(g)

From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.

$\therefore$ ${1 \over 2}$ moles of H₂(g) occupies = ${1 \over 2} \times 22.4$ = 11.2 L

$\therefore$ Option (A) is correct.

Option B :

According to avogadro Hypothesis, volume is directly proportional to the no of moles of gases.
Only for gases, no of mole and volume relationship is possible i.e, for two gases if no of moles are same then their volume is also same. But here HCL is in aqueous form and H₂ is in gaseous form so we can't find volume of HCL(aq) using no of moles of HCL(aq).

$\therefore$ Option (B) is wrong.

Option C :

2 moles of Al(s) produces = 3 moles of H₂(g)

$\therefore$ 1 mole of Al(s) produces = ${3 \over 2}$ moles of H₂(g)

From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.

$\therefore$ ${3 \over 2}$ moles of H₂(g) occupies = ${3 \over 2} \times 22.4$ = 33.6 L

6 moles of HCl(aq) produces = 3 moles of H₂(g)

$\therefore$ 1 mole of HCl(aq) produces = ${3 \over 6}$ moles of H₂(g) = ${1 \over 2}$ moles of H₂(g)

$\therefore$ ${1 \over 2}$ moles of H₂(g) occupies = ${1 \over 2} \times 22.4$ = 11.2 L

So from every mole of HCl(aq) that reacts only 11.2 L H₂(g) produces.

Form every mole of Al(s) produces 33.6 L H₂(g) but from every mole of HCl(aq) that reacts only 11.2 L H₂(g) produces not 33.6 L.

$\therefore$ Option (C) is wrong.

Option D :

2 moles of Al(s) produces = 3 moles of H₂(g)

$\therefore$ 1 mole of Al(s) produces = ${3 \over 2}$ moles of H₂(g)

From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.

$\therefore$ ${3 \over 2}$ moles of H₂(g) occupies = ${3 \over 2} \times 22.4$ = 33.6 L

$\therefore$ Option (D) is wrong.

2007 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2007 Paper 2 Offline

Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is

A.

3

B.

4

C.

5

D.

6

2006 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2006

How many moles of magnesium phosphate Mg₃(PO₄)₂ will contain 0.25 mole of oxygen atoms?

A.

1.25 $\times$ 10^-2

B.

2.5 $\times$ 10^-2

C.

0.02

D.

3.125 $\times$ 10^-2

2006 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2006

Density of a 2.05M solution of acetic acid in water is 1.02 g/mL The molality of the solution is

A.

2.28 mol kg^-1

B.

0.44 mol kg^-1

C.

1.14 mol kg^-1

D.

3.28 mol kg^-1

2005 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2005

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will

A.

be a function of the molecular mass of the substance

B.

remain unchanged

C.

increase two fold

D.

decrease twice

2005 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2005

Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture?

A.

2.70 M

B.

1.344 M

C.

1.50 M

D.

1.20 M

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is

A.

urea

B.

benzamide

C.

acetamide

D.

thiourea

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

6.02 $\times$ 10²⁰ molecules of urea are present in 100 ml of its solution. The concentration of urea solution is
(Avogadro constant, N_A = 6.02 $\times$ 10²³ mol^-1)

A.

0.02 M

B.

0.01 M

C.

0.001 M

D.

0.1 M

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H₃PO₃), the volume of 0.1 M aqueous KOH solution required is

A.

40 mL

B.

20 mL

C.

10 mL

D.

60 mL

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?

A.

67.2 L

B.

44.8 L

C.

22.4 L

D.

89.6 L

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was

A.

0.14

B.

0.28

C.

0.35

D.

0.07

2003 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2003 Screening

Mixture X = 0.02 mol of [Co(NH₃)₅SO₄]Br and 0.02 mol of [Co(NH₃)₅Br]SO₄ was prepared in 2 litre of solution.
1 litre of mixture X + excess AgNO₃ $ \to $ Y.
1 litre of mixture X + excess BaCl₂ $ \to $ Z
No. of moles of Y and Z are

A.

0.01, 0.01

B.

0.02, 0.01

C.

0.01, 0.02

D.

0.02, 0.02

2003 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2003 Screening

Which has maximum number of atoms?

A.

24 g of C(12)

B.

56g of Fe(56)

C.

27g of Al(27)

D.

108g of Ag(108)

Correct Answer: A

Explanation:

To determine which option has the maximum number of atoms, we will use Avogadro's number and the concept of moles. Avogadro's number ($6.02 \times 10^{23}$ atoms/mol) represents the number of atoms in one mole of any substance. The number of moles of a substance is calculated by dividing the given mass of the substance by its molar mass (which is given in grams per mole and is numerically equal to the element's atomic mass for pure elements).

Let's calculate the number of moles for each option:

Option A: 24 g of C (Carbon, atomic mass = 12 g/mol)

The number of moles of C = $\frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles}$

Therefore, the number of atoms in 24 g of C = $2 \text{ moles} \times 6.02 \times 10^{23} \text{ atoms/mol} = 1.204 \times 10^{24}$ atoms

Option B: 56 g of Fe (Iron, atomic mass = 56 g/mol)

The number of moles of Fe = $\frac{56 \text{ g}}{56 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 56 g of Fe = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Option C: 27 g of Al (Aluminum, atomic mass = 27 g/mol)

The number of moles of Al = $\frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 27 g of Al = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Option D: 108 g of Ag (Silver, atomic mass = 108 g/mol)

The number of moles of Ag = $\frac{108 \text{ g}}{108 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 108 g of Ag = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Comparing the number of atoms in each option, Option A with 1.204 $\times 10^{24}$ atoms has the maximum number of atoms.

2003 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2003

Calculate the molarity of water if it's density is 1000 kg/m³

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g mol^-1) is

A.

twice that in 60 g carbon

B.

6.023 $\times$ 10²²

C.

half that in 8 g He

D.

558.5 $\times$ 6.023 $\times$ 10²³

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

With increase of temperature, which of these changes?

A.

molality

B.

weight fraction of solute

C.

molarity

D.

mole fraction

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight . Molecular weight of the compound is 108 g mol^-1 . Molecular formula of compound is

A.

C₂H₆N₂

B.

C₃H₄N

C.

C₆H₈N₂

D.

C₉H₁₂N₃

Correct Answer: C

Explanation:

Element	Weight ratio	No of atoms of element	Simplest Ratio
C	9	9/12 = 3/4	3
H	1	1/1 = 1	4
N	3.5	3.5/14 = 1/4	1

Empherical Formula of the compound (EF) will be C₃H₄N

Now to find molecular formula at first find number of moles of given mass

No. of moles (n) =

Given mass

Molar mass

$ \Rightarrow $ ${{108} \over {54}}$

$ \Rightarrow $ 2

$\therefore$ No. of Moles is 2

Molecular formula (MF) = n (no. of moles) $\times$ Empherical Formula (EF)

$ \Rightarrow $ 2 $\times$ C₃H₄N

$ \Rightarrow $ C₆H₈N₂

$\therefore$ Option (C) is correct

2002 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2002 Screening

How many moles of electron weigh one kilogram?

A.

$6.023 \times {10^{23}}$

B.

${1 \over {9.108}} \times {10^{31}}$

C.

${{6.023} \over {9.108}} \times {10^{54}}$

D.

${1 \over {9.108 \times 6.023}} \times {10^8}$

Correct Answer: D

Explanation:

To find out how many moles of electrons weigh one kilogram, we need to understand the concept of the mole and the mass of an electron.

A mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions, electrons, etc.) as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's number, $N_A = 6.023 \times 10^{23}$ entities per mole.

The mass of an electron is approximately $9.109 \times 10^{-31}$ kg per electron.

The number of moles ($n$) is given by the equation:

$n = \frac{m}{M}$

where $m$ is the mass of the substance and $M$ is the molar mass of the substance.

In the case of electrons, the mass ($m$) we're considering is 1 kilogram, and we need to find the equivalent mass in terms of moles of electrons.

However, instead of using the molar mass as we would for atoms or molecules, we directly relate the mass of the electrons to Avogadro's number since we're dealing with elementary particles.

The calculation involves finding out how many electrons make up 1 kg, and then calculating how many moles of electrons that represents.

First, find the number of electrons in 1 kg:

$\text{Number of electrons} = \frac{1 \, \text{kg}}{(9.109 \times 10^{-31} \, \text{kg/electron})} = \frac{1}{9.109 \times 10^{-31}}$

Then, to find out how many moles this number of electrons represents, we divide by Avogadro's number:

$\text{Moles of electrons} = \frac{1}{9.109 \times 10^{-31}} \times \frac{1}{6.023 \times 10^{23}}$

Simplifying this gives us:

$\text{Moles of electrons} = \frac{1}{9.109} \times 10^{31} \times \frac{1}{6.023 \times 10^{23}} = \frac{1}{9.109 \times 6.023} \times 10^{8}$

Hence, the option that correctly represents the number of moles of electrons that weigh one kilogram is:

Option D: ${1 \over {9.108 \times 6.023}} \times {10^8}$

2001 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2001 Screening

In the standardization of Na₂S₂O₃ using K₂Cr₂O₇ by iodometry, the equivalent weight of K₂Cr₂O₇ is:

A.

(molecular weight)/2

B.

(molecular weight)/6

C.

(molecular weight)/3

D.

Same as molecular weight

2001 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2001 Screening

An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250ml. the volume of 0.1N NaOH required to completely neutralize 10ml of this solution is :

A.

40ml

B.

20ml

C.

10ml

D.

4ml

Correct Answer: A

Explanation:

To solve this problem, we need to use the concept of normality and the stoichiometry of the reaction between oxalic acid and sodium hydroxide. Oxalic acid dihydrate (C2H2O4.2H2O) reacts with sodium hydroxide (NaOH) in a 1:2 molar ratio because oxalic acid is a diprotic acid, meaning each molecule of oxalic acid can donate two protons.

First, we need to calculate the number of moles of oxalic acid in the solution:

The molar mass of oxalic acid dihydrate (C2H2O4.2H2O) is calculated as follows:

C: 12.01 g/mol * 2 = 24.02 g/mol

H: 1.01 g/mol * 4 = 4.04 g/mol

O: 16.00 g/mol * 6 = 96.00 g/mol

Total: 126.06 g/mol

Given mass of oxalic acid dihydrate = 6.3 g

Moles of oxalic acid dihydrate = $\frac{\text{mass}}{\text{molar mass}} = \frac{6.3 g}{126.06 g/mol} = 0.05 \text{ mol}$

Since the solution is made up to 250 mL, the molarity (M) of this solution is calculated as follows:

$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.05 \text{ mol}}{0.25 \text{ L}} = 0.2 \text{ M}$

When 10 mL of this solution reacts with NaOH, the volume of NaOH required to completely neutralize the solution can be found using the stoichiometry of the reaction:

The reaction between oxalic acid (H2C2O4) and NaOH is represented by the balanced chemical equation:

$H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$

This indicates that 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of oxalic acid in 10 mL of the solution can be found as:

$\text{Moles of oxalic acid} = 0.2 M * 0.01 L = 0.002 \text{ mol}$

Since each mole of oxalic acid requires 2 moles of NaOH for complete neutralization:

$\text{Moles of NaOH required} = 0.002 \text{ mol of oxalic acid} * 2 = 0.004 \text{ mol of NaOH}$

The normality (N) of NaOH solution is given as 0.1N. Because NaOH is monovalent, its normality is equal to its molarity in this context. Hence, the volume of NaOH solution required can be calculated using the formula:

$V = \frac{\text{moles of NaOH}}{\text{Normality of NaOH solution}} = \frac{0.004 \text{ mol}}{0.1 N} = 0.04 L = 40 mL$

Therefore, the correct answer is Option A: 40mL.

2001 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2001

Hydrogen peroxide solution (20 ml) reacts quantitatively with a solution of KMnO₄ solution is just decolourised by 10 ml of MnSO₄ in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO₂. The brown precipitated is dissolved in 10 ml of 0.2 M sodium oxalate under boiling condition in the presence of dilute H₂SO₄. Write the balanced equations involved in the reactions and calculate the molarity of H₂O₂.

Correct Answer: Molarity : 0.1

Explanation:

Let the molarity of $\mathrm{KMnO}_4$ be $M_1$. Moles of $\mathrm{KMnO}_4$ in $20 \mathrm{ml}$

$ =\frac{20 \times M_1}{1000} $

$n$-factor of $\mathrm{KMnO}_4$ when it reacts with $\mathrm{MnSO}_4$ is 3.

$\therefore$ Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{MnSO}_4=\frac{20 \times M_1}{1000} \times 3$

$\therefore$ Equivalent of $\mathrm{MnSO}_4$ reacting with $\mathrm{KMnO}_4=\frac{20 \times M_1}{1000} \times 3$

Since $\mathrm{MnSO}_4$ has $n$-factor 2, the mole of $\mathrm{MnSO}_4$ reacting

$ =\frac{20 \times M_1}{1000} \times \frac{3}{2} $

Total mole of $\mathrm{MnO}_2$ produced $=$ mole of $\mathrm{KMnO}_4+$ mole of $ \mathrm{MnSO}_4 $

Equivalent of $\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4$ reacting with $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$

$\therefore$ Equivalent of $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$

Mole of $\mathrm{MnO}_2$ reacting with sodium oxalate $=\frac{10 \times 0.2 \times 2}{1000 \times 2}$

[as $n$-factor for $\mathrm{MnO}_2$ is 2].

Therefore, $\frac{20 \times M_1}{1000}+\left[\frac{20 \times M_1}{1000} \times \frac{3}{2}\right]=\frac{10 \times 0.2}{1000} ; M_1=0.04$

Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{H}_2 \mathrm{O}_2=\frac{20 \times 0.04}{1000} \times 5$ $ =0.004 $

If molarity of $\mathrm{H}_2 \mathrm{O}_2$ is $M_2$, then $=\frac{20 \times M_2 \times 2}{1000}=0.004$

$\therefore M_2=0.1 \mathrm{M}$

The reactions involved are:

$2 \mathrm{KMnO}_4+5 \mathrm{H}_2\mathrm{O}_2+3 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{K}_2\mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2\mathrm{O}+5 \mathrm{O}_2$,

$2 \mathrm{KMnO}_4+3 \mathrm{MnSO}_4+2 \mathrm{H}_2\mathrm{O} \rightarrow 5 \mathrm{MnO}_2+2 \mathrm{H}_2\mathrm{SO}_4+\mathrm{K}_2\mathrm{SO}_4$,

$\mathrm{MnO}_2+\mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4+2 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{MnSO}_4+2 \mathrm{CO}_2+\mathrm{Na}_2\mathrm{SO}_4+2 \mathrm{H}_2\mathrm{O}$.

1999 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1999

How many millilitres of 0.5 M H₂SO₄ are needed to dissolve 0.5 g of copper(II) carbonate?

Correct Answer: 8.092 ml

Explanation:

To determine how many millilitres of 0.5 M $ \text{H}_2\text{SO}_4 $ are needed to dissolve 0.5 g of copper(II) carbonate $ (\text{CuCO}_3) $, we first need to write the chemical reaction between the copper(II) carbonate and sulfuric acid ( $ \text{H}_2\text{SO}_4 $ ):

$ \text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2 $

From this reaction, it's observable that 1 mole of copper(II) carbonate reacts with 1 mole of sulfuric acid to produce 1 mole of copper(II) sulfate, water, and carbon dioxide.

Let's calculate the number of moles of copper(II) carbonate in 0.5 g:

The molar mass of $ \text{CuCO}_3 $ is calculated as:

$ \text{Molar mass of } \text{CuCO}_3 = (\text{Molar mass of Cu}) + (\text{Molar mass of C}) + 3 \times (\text{Molar mass of O}) $

$ \text{Molar mass of } \text{CuCO}_3 = (63.55 \text{ g/mol}) + (12.01 \text{ g/mol}) + 3 \times (16.00\text{ g/mol}) $

$ \text{Molar mass of } \text{CuCO}_3 = 63.55 + 12.01 + 48.00 $

$ \text{Molar mass of } \text{CuCO}_3 = 123.56 \text{ g/mol} $

Now, we'll find out the number of moles in 0.5 g of $ \text{CuCO}_3 $:

$ \text{Number of moles of } \text{CuCO}_3 = \frac{\text{mass of } \text{CuCO}_3}{\text{Molar mass of } \text{CuCO}_3} $

$ \text{Number of moles of } \text{CuCO}_3 = \frac{0.5 \text{ g}}{123.56 \text{ g/mol}} $

$ \text{Number of moles of } \text{CuCO}_3 \approx 0.004046 \text{ mol} $

Since the reaction is a 1:1 molar ratio, the moles of sulfuric acid needed would also be $ 0.004046 \text{ mol} $.

The concentration (C) of a solution is given by the formula:

$ C = \frac{n}{V} $

where:

$ C $ is the concentration of the solution in moles per liter (M).

$ n $ is the number of moles of solute.

$ V $ is the volume of solution in liters.

Rearranging the formula to solve for volume $ V $:

$ V = \frac{n}{C} $

The given concentration of $ \text{H}_2\text{SO}_4 $ is 0.5 M, which means 0.5 moles of $ \text{H}_2\text{SO}_4 $ per liter. Now, let's calculate the volume required for the $ 0.004046 \text{ mol} $ of $ \text{H}_2\text{SO}_4 $:

$ V = \frac{0.004046 \text{ mol}}{0.5 \text{ M}} = \frac{0.004046 \text{ mol}}{0.5 \text{ mol/L}} $

$ V = 0.008092 \text{ L} $

To convert liters to millilitres:

$ V(\text{mL}) = 0.008092 \text{ L} \times 1000 \frac{\text{mL}}{\text{L}} $

$ V(\text{mL}) = 8.092 \text{ mL} $

Therefore, you will need approximately 8.092 mL of 0.5 M $ \text{H}_2\text{SO}_4 $ to completely dissolve 0.5 g of copper(II) carbonate under the assumption that the reaction goes to completion with no side reactions.

1999 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1999

A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm³/g. If the virus is considered to be a single particle, find its molar mass.

1998 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1998

An aqueous solution containing 0.10 g KIO₃ (formula weight = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I₂ consumed 45.0 mL of thiosulphate solution to decolourise the blue strach-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Correct Answer: 0.0623

Explanation:

To calculate the molarity of the sodium thiosulphate solution, we'll need to apply stoichiometry principles. First, we must determine the number of moles of $ \text{KIO}_3 $ as it will relate to the number of moles of $ \text{I}_2 $ liberated during the reaction. Then we will use the volume of the sodium thiosulphate solution to find its molarity.

The reaction for the iodometry where $ \text{I}_2 $ is liberated from $ \text{KIO}_3 $ and reacted with the excess of $ \text{KI} $ is:

$ \text{KIO}_3 + 5 \text{KI} + 6 \text{HCl} \rightarrow 3 \text{I}_2 + 6 \text{KCl} + 3 \text{H}_2\text{O} $

From the balanced equation, we see that 1 mole of $ \text{KIO}_3 $ produces 3 moles of $ \text{I}_2 $.

Now let's find the number of moles of $ \text{KIO}_3 $:

$ \text{Moles of } \text{KIO}_3 = \frac{\text{mass of } \text{KIO}_3}{\text{molar mass of } \text{KIO}_3} $

$ = \frac{0.10 \text{ g}}{214.0 \text{ g/mol}} $

To get the number of moles, we need to perform the division:

$ = \frac{0.10}{214.0} $

$ = 4.6728972 \times 10^{-4} \text{ moles of } \text{KIO}_3 $

Since 1 mole of $ \text{KIO}_3 $ yields 3 moles of $ \text{I}_2 $, we multiply the moles of $ \text{KIO}_3 $ by 3:

$ \text{Moles of } \text{I}_2 = 3 \times 4.6728972 \times 10^{-4} $

$ = 1.4018692 \times 10^{-3} \text{ moles of } \text{I}_2 $

Next, $ \text{I}_2 $ reacts with sodium thiosulphate ($ \text{Na}_2\text{S}_2\text{O}_3 $) in the following stoichiometric reaction:

$ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 $

From the reaction, 1 mole of $ \text{I}_2 $ reacts with 2 moles of sodium thiosulphate. Therefore, the moles of sodium thiosulphate required to react with the $ \text{I}_2 $ produced can be calculated as follows:

$ \text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = 2 \times \text{Moles of } \text{I}_2 $

$ = 2 \times 1.4018692 \times 10^{-3} $

$ = 2.8037384 \times 10^{-3} \text{ moles of } \text{Na}_2\text{S}_2\text{O}_3 $

Now we have to calculate the molarity of the sodium thiosulphate solution. Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:

$ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} $

We have the number of moles of sodium thiosulphate and the volume is given as 45.0 mL which needs to be converted to liters:

$ \text{Volume in liters} = 45.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.045 \text{ L} $

Substitute the number of moles and the volume into the molarity formula:

$ M = \frac{2.8037384 \times 10^{-3} \text{ moles}}{0.045 \text{ L}} $

Performing the division gives us the molarity:

$ M = \frac{2.8037384 \times 10^{-3}}{0.045} $

$ = 0.0623053 \text{ M} $

So, the molarity of the sodium thiosulphate solution is approximately $ 0.0623 $ M.

1996 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1996

A 3.00 g sample containing Fe₃O₄, Fe₂O₃ and an inert impure substance is treated with excess of KI solution in presence of dilute H₂SO₄. The entire iron is converted into Fe²⁺ along with the liberation of iodine. The resulting solution is diluted to 100 ml, A 20 ml of the diluted solution requires 11.0 ml of 0.5 M Na₂S₂O₃ solution to reduce the iodine present. A 50 ml of the diluted solution, after complete extraction of the iodine requires 12.80 ml of 0.25 M KMnO₄ solution in the dilute H₂SO₄ medium for the oxidation of Fe²⁺. Calculate the percentages of Fe₂O₃ and Fe₃O₄ in the original sample.

Correct Answer: 49.33%, 34.8%

Explanation:

Let the original sample contains $x$ millimol of $\mathrm{Fe}_3 \mathrm{O}_4$ and $y$ millimol of $\mathrm{Fe}_2 \mathrm{O}_3$. In the first phase of reaction,

$ \begin{aligned} & \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{I}^{-} \longrightarrow 3 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_3 \mathrm{O}_4=2\right) \\\\ & \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_2 \mathrm{O}_3=2\right) \end{aligned} $

$\begin{aligned} \Rightarrow \text { Meq of } \mathrm{I}_2 \text { formed } & =\mathrm{Meq}\left(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{Fe}_2 \mathrm{O}_3\right) =\text { Meq of hypo required } \\\\ \Rightarrow 2 x+2 y & =11 \times 0.5 \times 5=27.5 ...........(i)\end{aligned}$

Now, total millimol of $\mathrm{Fe}^{2+}$ formed $=3 x+2 y$. In the reaction

$ \begin{array}{ll} \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+} \\\\ n \text {-factor of } \mathrm{Fe}^{2+}=1 \\\\ \Rightarrow \text { Meq of } \mathrm{MnO}_4^{-}=\text {Meq of } \mathrm{Fe}^{2+} \\\\ \Rightarrow 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32 ...........(ii) \end{array} $

Solving Eqs. (i) and (ii), we get

$ \begin{aligned} x=4.5 \text { and } y=9.25 \end{aligned} $

$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{4.5}{1000} \times 232=1.044 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{1.044}{3} \times 100=34.80 \% \\\\ \text { Mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{9.25}{1000} \times 160=1.48 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{1.48}{3} \times 100=49.33 \%\end{aligned}$

1994 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1994

8.0575 $\times$ 10^-2 kg of Glauber's salt is dissolved in water to obtain 1 dm³ of a solution of density 1077.2 kg.m^-3. Calculate the molarity, molality and mole fraction Na₂SO₄ in the solution.

Correct Answer: 0.25 M, 0.24 m, 4.3 $$\times$$ 10-3

Explanation:

Glauber's salt is $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}$.

Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4=142$

Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}=322$

Weight of Glauber's salt taken $=8.0575 \times 10^{-2} \mathrm{~kg}=80.575 \mathrm{~g}$

Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $322 \mathrm{~g}$ of Glauber's salt

$=142 / 322 \mathrm{~g}$

$\therefore$ Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $80.575 \mathrm{~g}$ of Glauber's salt

$ =\frac{142}{322} \times 80.575=35.53 \mathrm{~g} $

Hence number of moles of $\mathrm{Na}_2 \mathrm{SO}_4$ per $\mathrm{dm}^3$ of solution

$ =\frac{35.53}{142}=0.25 $

So, the molarity of solution is $0.25 \mathrm{M}$

$ \begin{aligned} \text { Density of solution } & =1077.2 \mathrm{~kg} \mathrm{~m}^{-3} \\\\ & =\frac{1077.2 \times 10^3}{10^6} \mathrm{~g} \mathrm{~cm}^{-3}=1.0772 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned} $

$\begin{aligned} & \text { Weight of solution }=\text { volume } \times \text { density } \\\\ & =1000 \times 1.0772 \mathrm{~g}=1077.2 \mathrm{~g} \\\\ & \text { Weight of water }=(1077.2-35.53)=1041.67 \mathrm{~g} \\\\ & \text { Molality of solution }=\frac{0.25}{1041.67} \times 1000=0.24 \mathrm{~m}\end{aligned}$

$\begin{aligned} & \text { Number of moles of water in solution }=\frac{1041.67}{18}=57.87 \\\\ & \text { Mole fraction of } \mathrm{Na}_2 \mathrm{SO}_4=\frac{\text { Moles of } \mathrm{Na}_2 \mathrm{SO}_4}{\text { Total number of moles }} \\\\ & \qquad=\frac{0.25}{0.25+57.87}=0.0043 \text { or } 4.3 \times 10^{-3}\end{aligned}$

1994 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1994

The composition of a sample of Wustite is Fe_0.93O_1.00 what percentage of the iron is present in the form of Fe(III)?

Correct Answer: 15.05%

Explanation:

To determine the percentage of iron present as Fe(III) in the sample of wustite (Fe_0.93O_1.00), we need to assume that the rest of the iron that is not Fe(II) is present as Fe(III). This is because wustite, which nominally has the formula FeO, has a non-stoichiometric composition due to the presence of both Fe(II) and Fe(III) ions that maintain charge neutrality in the lattice.

Given the formula Fe_0.93O_1.00, for every one oxygen atom, there are 0.93 irons. In terms of charge, the oxygen anion has a charge of -2. The formula can be interpreted as having 0.93 moles of Fe per mole of wustite, and we need to balance the charges considering the presence of both Fe(II) and Fe(III).

Fe(II) has a charge of +2, and Fe(III) has a charge of +3. If we let x be the fraction of Fe(III), then the fraction of Fe(II) would be 0.93 - x because the total amount of iron is 0.93 moles of iron (Fe). Now, we can set up a charge balance equation:

Charge from Fe(II) + Charge from Fe(III) = Charge from O

(0.93 - x)(+2) + (x)(+3) = 1(+2)

2(0.93 - x) + 3x = 2

1.86 - 2x + 3x = 2

Combining like terms gives:

1.86 + x = 2

x = 2 - 1.86

x = 0.14

So, 0.14 moles of Fe are in the form of Fe(III) per mole of wustite.

To find the percentage of Fe(III), you divide the number of moles of Fe(III) by the total moles of Fe and multiply by 100%:

$ \text{Percentage of Fe(III)} = \left( \frac{x}{0.93} \right) \times 100\% $

$ \text{Percentage of Fe(III)} = \left( \frac{0.14}{0.93} \right) \times 100\% $

$ \text{Percentage of Fe(III)} = (0.1505) \times 100\% $

$ \text{Percentage of Fe(III)} = 15.05\% $

Therefore, 15.05% of the iron is present in the form of Fe(III) in the sample of wustite Fe_0.93O_1.00.

1994 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1994

The compound YBa₂Cu₃O₇, which shows superconductivity, has copper in oxidation state ______ assume that the rare earth element yttrium is in its usual +3 oxidation state.

Chemistry Chapters

Some Basic Concepts of Chemistry

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